Chern-Simons term in Coulomb or radiation gauge

Yes.

The question now is to show $$ int d^2xepsilon^{munusigma}a_{mu} partial_{mu}a_{sigma}=int d^2x [2 a_0(partial_1 a_2-partial_2 a_1)+(a_2partial_0 a_1-a_1partial_0 a_2)] =int d^2x 2 a_0(partial_1 a_2-partial_2 a_1) tag{1}. $$ That is, $$ int d^2x (a_2partial_0 a_1-a_1partial_0 a_2)=int d^2x vec{a}times partial_0 vec{a}=0 tag{2}. $$ The Coulomb gauge condition ($nabla cdot vec{a}=0$) in momentum space is $vec{k}cdotvec{a}(vec{k})=0$. Differentiating with respect to time and $krightarrow -k$ gives $vec{k}cdot partial_0 vec{a}(-vec{k})=0.$. These two equations imply that $vec{a}(vec{k})$ and $partial_0 vec{a}(-vec{k})$ are both perpendicular to $vec{k}$, which in turn imply that they are parallel to each other: $$ vec{a}(vec{k}) times partial_0 vec{a}(vec{-k})=0 tag{3}. $$ Now, representing (2) in momentum space gives $$ int d^2x vec{a}times partial_0 vec{a}=int frac{d^2k}{(2pi)^2}vec{a}(vec{k})times partial_0 vec{a}(-vec k)=0,tag{4} $$ and the claim is proved.